Why is this function injective?

[Wiz14]

Homework Statement

The function from R to R satisfies x + f(x) = f(f(x)) Find all Solutions of the equation f(f(x)) = 0.

Part of the problem solution says that if f(x) = f(y), then "obviously" x = y. I understand the rest of the solution, but why does f(x) = f(y) imply that x = y?

 

[Hypersphere]

Have you tried comparing f(f(x)) and f(f(y))?

 

[Wiz14]

Have you tried comparing f(f(x)) and f(f(y))?

sorry I am new to this stuff, but do you mean f(f(x)) = 0 implies f(f(y)) = 0 which implies f(f(x)) = f(f(y)) ? If yes then how does this prove f(x) = f(y)?

 

[HallsofIvy]

No, he meant nothing of the sort. And f(f(x))= 0 does NOT imply f(f(y))= 0.

 

[Wiz14]

No, he meant nothing of the sort. And f(f(x))= 0 does NOT imply f(f(y))= 0.

If f(f(x)) = 0, independently of the argument f(x), then doesn't substituting f(y) for f(x) give f(f(y)) = 0?

 

[Office_Shredder]

but why does f(x) = f(y) imply that x = y?

If f(x) = f(y), then f(f(x)) = f(f(y)). Use this to prove that x=y. This has nothing to do with anything equaling zero

 

[Wiz14]

If f(x) = f(y), then f(f(x)) = f(f(y)). Use this to prove that x=y. This has nothing to do with anything equaling zero

If f(y) = f(x) then substituting f(y) into the original equation gives f(f(y) = f(x) + y = f(x) + x, then subtracting the f(x) from the last equation gives x = y, is this correct? Thanks for the help.

 

[Hypersphere]

If f(y) = f(x) then substituting f(y) into the original equation gives f(f(y) = f(x) + y = f(x) + x, then subtracting the f(x) from the last equation gives x = y, is this correct? Thanks for the help.

That is right.